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ser-zykov [4K]
4 years ago
11

The correct simplification of the quadratic formula used to solve 8x 2 - 2x = 1?

Mathematics
2 answers:
suter [353]4 years ago
6 0

Answer:

x_{1}=\frac{1}{2}\\x_{2}=-\frac{1}{4}

Step-by-step explanation:

Hello

For ax2 + bx + c = 0,the values of x which are the solutions of the equation are given by

x=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

± means  there are two solution for x,  X1 and X2

x_{1} =\frac{-b+\sqrt{b^{2}-4ac } }{2a} \\x_{2} =\frac{-b-\sqrt{b^{2}-4ac } }{2a}

Step 1

convert 8x2-2x=1 into the form ax2 + bx + c = 0 ( right side equal to cero)

subtract 1 in each side

8x^{2} -2x=1\\8x^{2} -2x-1=1-1\\8x^{2} -2x-1=0

Step 2

replace in the equation

Let

a=8

b=-2

c=-1

Hence

x_{1} =\frac{-b+\sqrt{b^{2}-4ac }}{2a}\\ x_{1}=\frac{-(-2)+\sqrt{(-2)^{2}-4(8)(-1) } }{2*8}\\x_{1} =\frac{+2+\sqrt{4+32 }}{16}\\x_{1} =\frac{+2+\sqrt{36 } }{16}\\x_{1} =\frac{2+\sqrt{36 }}{16}\\x_{1} =\frac{2+6}{16} \\x_{1} =\frac{8}{16} \\x_{1}=-\frac{1}{2}

Now, X2

x_{2} =\frac{-b-\sqrt{b^{2}-4ac }}{2a}\\ x_{2}=\frac{-(-2)-\sqrt{(-2)^{2}-4(8)(-1) } }{2*8}\\x_{1} =\frac{+2-\sqrt{4+32 }}{16}\\x_{2} =\frac{+2-\sqrt{36 } }{16}\\x_{2} =\frac{2-\sqrt{36 }}{16}\\x_{2} =\frac{2-6}{16} \\x_{2} =\frac{-4}{16} \\x_{2}=-\frac{1}{4}

Have a great day

Phantasy [73]4 years ago
4 0
<span>(2x−1)</span><span>(4x+1<span>)..................hope this helps

</span></span>
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